Current Electricity question for JEE Main (JEE Main 2020 (5 Sep, Shift 2)), with solution

Answer: (D) $+1$ V

In the figure the longer plate is the positive terminal.

**Branch DC:** at junction C, $1$ A arrives from A and $2$ A leaves towards F, so $1$ A flows from D to C through the $2\ \Omega$ resistor:
$$V_D - V_C = 1\times2 = 2\ \text{V}$$

**1 V cell (A to C):** its positive plate faces C, so $V_C - V_A = +1$ V.

**2 V cell (D to B):** its positive plate faces D, so $V_B - V_D = -2$ V.

Going from A to B via C and D:
$$V_B - V_A = (V_C - V_A) + (V_D - V_C) + (V_B - V_D) = 1 + 2 - 2 = +1\ \text{V}$$

Solution by Sreeraj P, M.Sc Physics